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How do I write the Laurent series expansion of $\frac{1}{z-3}$ for $|z-3|>5$.

I'm trying to rewrite this into the geometric series form of $\frac{1}{1-r}$. I rewrote it as $-\frac{1}{3}(\frac{1}{1-\frac{z}{3}})$, but I don't think this is correct. I'm pretty sure that because the disk is centered around 3 I need to keep a $z-3$ in the "r" spot.

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  • $\begingroup$ Not sure what you’re confused about, and also not sure why is there an arbitrary-seeming “5” there, when the Laurent series is valid for $|z-3| > 0$. Anyway, the Laurent series can have negative exponents, so the Laurent series is itself, i.e. $1 /( z - 3)$. Edit: if you’re looking to expand about a different point (not 3), then ignore the above, but you should clarify that in your question. $\endgroup$ Jul 15 '20 at 16:45
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    $\begingroup$ take the disk into consideration. its a disk of center 3 and radius 5, and you're working outside that disk, so you know (z-3)/5 is bigger then 1, you can use that as your reference for $r$ $\endgroup$
    – Rye
    Jul 15 '20 at 18:29
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I made a mistake last night as honestly as possible, I hope below formulas can help you $$|z_1-z_2+z_2|\leq|z_1-z_2|+|z_2|\\|z_1-z_2|\geq |z_1|-|z_2|\\|z-2|\geq |z_-3|-|1|\\|z-2|+1\geq |z-3|>5 \\ \to |z-2|>4$$ you need soomething converge to $\frac1{z-3}$ if we look at $\frac r{1-r}=r+r^2+r^3+...$ we have to find $r$ such that series converge to $\frac1{z-3}$so $$\frac{r}{1-r}=\frac{1}{z-3}\to rz-3r=1-r\\r(z-3+1)=1\to r=\frac{1}{z-3+1}=\frac{1}{z-2}$$ now lets look at $|z-2|>4$ so $\frac{1}{|z-2|}<\frac 14$ convergence condition
finally we have $$\frac1{z-3}=\frac{r}{1-r}=\frac{\frac{1}{z-2}}{1-\frac{1}{z-2}}=(\frac{1}{z-2})+(\frac{1}{z-2})^2+(\frac{1}{z-2})^3+...$$

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  • $\begingroup$ The series in the last formula is not a Laurent series and doesn't converge to $1/(z - 3)$. $\endgroup$
    – Maxim
    Jul 16 '20 at 12:46
  • $\begingroup$ @Maxim: Hi, please take a look at the complete editing. is it ok now? $\endgroup$
    – Khosrotash
    Jul 16 '20 at 14:29
  • $\begingroup$ We have $\sum_{k \geq 1} (z - 2)^{-k} = 1/(z - 3)$ for $|z - 2| > 1$, which covers the domain $|z - 3| > 5$. So formally one can say that this is a valid answer. Typically though $|z - z_0| > R$ means we're considering a series centered at $z = z_0$. Then the answer is just $1/(z - 3)$. $\endgroup$
    – Maxim
    Jul 16 '20 at 14:59

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